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A tennis ballis dropped on to the floor ...

A tennis ballis dropped on to the floor from a height of 4m. It rebounds to a height of 2m. If the ball is in contact with the floor for 1`2xx10^(-3)s`, its average acceleration during

A

0

B

`1260ms^(-2)`

C

`980ms^(-2)`

D

`600ms^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`V^(2)=u^(2)+2as`
As u=0 , hence `V^(2)= 2as` or `V = sqrt2as`
When the tennis ball is dropped the velocity gained by it just before colliding with the floor is :
`u= sqrt2xx9.8xx4 = 8.85ms^(-1)` (upwords) `=- 8.85ms^(-1)` (upwords)
time of contact of the ball with the floor
(Given `t = 12xx10^(-3)S`
Hence , average acceleration
`a= (v-u)/(t) = (6.26-(-8.85)/(12xx10^(-3)`
`=+1260ms^(-2)`
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